How do you factor the trinomial #5x^2+ 6x + 1#?

Question

Avery Alexander · Accepted Answer

#5x^2+6x+1=(5x+1)(x+1)#

#5x^2+6x+1# is a quadratic equation, #ax^2+bx+c#, where #a=5, b=6, and c=1#.

Use the AC method.

Multiply #a# times #c#.

#5xx1=5#

Determine which numbers when added equal #6# and when multiplied equal #5#. The numbers #1# and #5# meet the criteria.

Rewrite the trinomial with #5x# and #x# in place of #6x#.

#5x^2+5x+x+1#

Group the first two terms and the last two terms.

#(5x^2+5x)+(x+1)#

Factor out #5x# from the first group.

#5x(x+1)+1(x+1)# (I included the #1# in front of the parentheses so you can see where the final answer comes from.)

Factor out #(x+1)#.

#(5x+1)(x+1)#

Eva Adamson · Answer

undefined To factor the trinomial $5x^2 + 6x + 1$, you can use the factoring by grouping method. Start by multiplying the coefficient of the quadratic term (5) by the constant term (1), which gives 5. Then find two numbers that multiply to give 5 and add to give the coefficient of the linear term (6). These numbers are 5 and 1. Split the middle term using these numbers: $5x^2 + 5x + x + 1$. Then factor by grouping: $5x(x + 1) + 1(x + 1)$. Now, factor out the common factor $x + 1$: $(5x + 1)(x + 1)$. So, the factored form of $5x^2 + 6x + 1$ is $(5x + 1)(x + 1)$.

Eva Adamson · Answer

undefined To factor the trinomial $5x^2 + 6x + 1$, you can use the factoring by grouping method or the quadratic formula. However, this particular trinomial doesn't factor nicely into linear factors with integer coefficients. So, if you're asked to factor it, you would write it as $ (5x + 1)(x + 1) $.